xy + x - 2y + 1 = 0
xy + x - 2y - 2 = -3
(xy - 2y) + (x - 2) = -3
y(x - 2) + (x - 2) = -3
(x - 2)(y + 1) = -3
Vì x, y nguyên \(\Rightarrow\) x - 2 \(\in\)Ư(-3) = {1; -1; 3; -3}
Lập bảng giá trị
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
| y + 1 | -3 | 3 | -1 | 1 |
| y | -4 | 2 | -2 | 0 |
Vậy (x; y) \(\in\) { (3; -4) , (1; 2) , (5; -2) , (-1; 0) }
\(xy+x-2y+1=0\)
\(\Leftrightarrow x\left(y+1\right)-2y+1=0\)
\(\Leftrightarrow x\left(y+1\right)-2y-2=3\)
\(\Leftrightarrow x\left(y+1\right)-2\left(y+1\right)=3\)
\(\Leftrightarrow\left(y+1\right)\left(x-2\right)\inƯ\left(3\right)\)
Ta có các trường hợp :
+) \(\left\{{}\begin{matrix}y+1=1\\x-2=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=5\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}x-2=1\\y+1=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}x-2=-1\\y+1=-3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-4\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}x-2=-3\\y+1=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)
Vậy....
